I’m graphing transformations of the cotangent function, let’s graph y equals negative cotangent of theta plus pi over 2. Now let’s start by recalling some key points of the cotangent function and I’ll use u as sort of a dummy variable here, cotangent of u.

Now we’ve plotted 0 pi over 4 pi over 2, 3 pi over 4 and pi previously and we found that at 0 and pi cotangent was undefined and pi over 2 cotangent was 0. Pi over 4 cotangent was 1 and 3 pi over 4 -1. So these indicates asymptotes, vertical asymptotes at 0 and pi but what happens when we have this transformation minus cotangent of theta plus pi over 2?

Well I'm going to have to make oops, cotangent, let’s make a little substitution. Let’s call theta plus pi over 2 u. And so if u equals theta plus pi over 2 then u minus pi over 2 equals theta. So all I have to do to get the theta value is from my graph is subtract pi over 2 from these guys. So let me do that right now, 0 minus pi over 2 negative pi over 2, pi over 4 minus pi over 2, negative pi over 4, 0 pi over 4 pi over 2. Now what do I do with cotangent u, I take its opposite like. This is cotangent if u the values I have here. I take the opposite of those values and those are going to be my new values and the opposite of undefined is still undefined but this becomes negative 1, 0, 1 undefined and so I’m ready to plot one period of my cotangent function.

Let me start by plotting the asymptotes, the vertical asymptotes at negative pi over 2 and pi over 2. So that’s here and here. Then let me plot my three points; negative pi over 4 negative 1, 0,0 and pi over 4,1. Negative pi over 4 is right between negative pi over 2 and 0 so negative pi over 4,1 is right, -1 is right here, 0,0 is here and pi over 4 1 is here. So it looks like because of this negative sign we have a reflection across the x axis. Cotangent usually decreases but this one is going to actually increase so it will look like this.

And remember if you want to graph more periods of your cotangent function all you have to do is duplicate this period shift it to the right, so if I do that this asymptote shifted one period to the right is going to be x equals 2 pi over 2. Let me draw that, and then I shift the three points, the three key points, these three. This one shifted to the right pi becomes 0, this one becomes 3 pi over 4 -1 and this one becomes 5 pi over 4, 1. And so I’m just graphing that and we can shift to the left as well and that will give us three nice periods of this cotangent function.

Right and this is also a vertical asymptote here that -3 pi over 2. When you look at this thing, this graph what you realize is this is exactly the tangent function. The tangent function does this, it’s got asymptotes at negative pi over 2 and pi over 2, it increases passes through 0,0, this is y equals tangent theta. And so what that means is, tangent theta is a reflection of cotangent shifted pi over 2 to the left, cotangent theta reflected across the x axis. Flip this way and then shifted to the left pi over 2, that gave us tangent.

That’s important. It's important for you to know that cotangent and tangent basically have the same shape graph, it’s just that all you have to do is reflect across the x axis and shift the right or left to get the other.